Section 11.4 The Comparison Tests
46. For what values of p does the series
∞
P
n=2 1
(npln n) converge?
Solution:
SECTION 11.4 THE COMPARISON TESTS ¤ 141 32. Use the Limit Comparison Test with = 1
1+1 and = 1
. lim
→∞
= lim
→∞
1+1 = lim
→∞
1
1 = 1
since lim
→∞1= 1by l’Hospital’s Rule , so ∞
=1
1
diverges [harmonic series] ⇒ ∞
=1
1
1+1 diverges.
33.
10
=1
1
5 + 5 = 1
5 + 15 + 1
5 + 25 + 1
5 + 35 + · · · + 1
5 + 105 ≈ 019926. Now 1
5 + 5 1
5, so the error is
10≤ 10≤
∞ 10
1
5 = lim
→∞
10
−5 = lim
→∞
−1 44
10
= lim
→∞
−1 44 + 1
40,000
= 1
40,000 = 0000 025.
34.
10
=1
1
4 = 11 14 +12
24 + 13
34 + · · · +110
104 ≈ 284748. Now1
4 ≤
4 for ≥ 1, so the error is
10≤ 10≤
∞ 10
4 = lim
→∞
10
−4 = lim
→∞
−
33
10= lim
→∞
−
33 + 3000
=
3000 ≈ 0000 906.
35. 10
=1
5−cos2 = cos21
5 +cos22
52 + cos23
53 + · · · + cos210
510 ≈ 007393. Now cos2 5 ≤ 1
5, so the error is
10≤ 10≤
∞
10
1
5 = lim
→∞
10
5− = lim
→∞
−5−
ln 5
10
= lim
→∞
−5−
ln 5+5−10 ln 5
= 1
510ln 5 64 × 10−8.
36. 10
=1
sin2
3 = sin21
1 +sin22
8 + sin23
27 + · · · +sin210
1000 ≈ 083253. Nowsin2
3 ≤ 1
3, so the error is
10≤ 10≤
∞
10
1
3 = lim
→∞
− 1 22
10
= lim
→∞
− 1 22 + 1
200
= 1
200 = 0005.
37. Since
10 ≤ 9
10 for each , and since ∞
=1
9
10 is a convergent geometric series
|| = 101 1, 0123 = ∞
=1
10 will always converge by the Comparison Test.
38. Clearly, if 0 then the series diverges, since lim
→∞
1
ln = ∞. If 0 ≤ ≤ 1, then ln ≤ ln ⇒ 1
ln ≥ 1
ln and ∞
=2
1
ln diverges (Exercise 11.3.21), so ∞
=2
1
ln diverges. If 1, use the Limit Comparison Test with = 1
ln and = 1
. ∞
=2
converges, and lim
→∞
= lim
→∞
1
ln = 0, so ∞
=2
1
ln also converges.
(Or use the Comparison Test, since ln for .) In summary, the series converges if and only if 1.
39. Since
converges, lim
→∞= 0, so there exists such that |− 0| 1 for all ⇒ 0 ≤ 1for all ⇒ 0 ≤ 2≤ . Since
converges, so does
2by the Comparison Test.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
47. Show that
∞
X
n=2
1 (ln n)ln ln n
diverges. [Hint: Use Formula 1..5.10(xr= er ln x) and the fact that ln x <√
x for x ≥ 1.]
Solution:
1088 ¤ CHAPTER 11 SEQUENCES, SERIES, AND POWER SERIES 46. Clearly, if 0 then the series diverges, since lim
→∞
1
ln = ∞. If 0 ≤ ≤ 1, then ln ≤ ln ⇒ 1
ln ≥ 1
ln and ∞
=2
1
ln diverges (Exercise 11.3.23), so ∞
=2
1
ln diverges. If 1, use the Limit Comparison Test with = 1
ln and = 1
. ∞
=2
converges, and lim
→∞
= lim
→∞
1
ln = 0, so ∞
=2
1
ln also converges.
(Or use the Comparison Test, since ln for .) In summary, the series converges if and only if 1.
47. Using Formula 1.5.10, we have (ln )ln ln = (ln ln )(ln ln )
[ 1] = (ln ln )2 (
√ln )2 = ln = . So 1
(ln )ln ln 1
for all ≥ 2. Thus, the series ∞
=2
1
(ln )ln ln diverges by comparison with the partial harmonic series
∞
=2
1
, which diverges.
48. (a) Since lim
→∞() = 0, there is a number 0 such that |− 0| 1 for all , and so since
and are positive. Thus, since
converges, so does
by the Direct Comparison Test.
(b) (i) If = ln
3 and = 1
2, then lim
→∞
= lim
→∞
ln
= lim
→∞
ln
= limH
→∞
1
1 = 0. Now
is a convergent
series [ = 2 1], so ∞
=1
ln
3 converges by part (a).
(ii) If = 1 − cos
1
2
and = 1
2, then
lim→∞
= lim
→∞
1 − cos(12) 12 = lim
→∞
1 − cos(12) 12
= limH
→∞
sin(12) · (−23)
−23 = lim
→∞sin
1
2
= 0.
Now
is a convergent series [ = 2 1], so ∞
=1
1 − cos 1
2
converges by part (a).
49. (a) Since lim
→∞
= ∞, there is an integer such that
1whenever . (Take = 1 in Definition 11.1.3.) Then whenever and since
is divergent,
is also divergent by the Comparison Test.
(b) (i) If = 1
ln and = 1
for ≥ 2, then lim
→∞
= lim
→∞
ln = lim
→∞
ln
= limH
→∞
1
1 = lim
→∞ = ∞, so by part (a), ∞
=2
1
ln is divergent.
(ii) If = ln
and = 1
, then ∞
=1
is the divergent harmonic series and lim
→∞
= lim
→∞ln = lim
→∞ln = ∞, so ∞
=1
diverges by part (a).
50. Let = 1
2 and = 1
. Then lim
→∞
= lim
→∞
1
= 0, but
diverges while
converges.
° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
52. Show that if an > 0 andP an is convergent, thenP ln(1 + an) is convergent.
Solution:
142 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 40. (a) Since lim
→∞() = 0, there is a number 0 such that |− 0| 1 for all , and so since
and are positive. Thus, since
converges, so does
by the Comparison Test.
(b) (i) If = ln
3 and = 1
2, then lim
→∞
= lim
→∞
ln
= lim
→∞
ln
= limH
→∞
1
1 = 0, so ∞
=1
ln
3 converges by part (a).
(ii) If = ln
√ and = 1
, then lim
→∞
= lim
→∞
ln √
= lim
→∞
ln √
= limH
→∞
1
1(2√
) = lim
→∞
√2
= 0. Now
is a convergent geometric series with ratio = 1 [|| 1], so
converges by part (a).
41. (a) Since lim
→∞
= ∞, there is an integer such that
1whenever . (Take = 1 in Definition 11.1.5.) Then whenever and since
is divergent,
is also divergent by the Comparison Test.
(b) (i) If = 1
ln and = 1
for ≥ 2, then lim
→∞
= lim
→∞
ln = lim
→∞
ln
= limH
→∞
1
1 = lim
→∞ = ∞, so by part (a), ∞
=2
1
ln is divergent.
(ii) If = ln
and = 1
, then ∞
=1
is the divergent harmonic series and lim
→∞
= lim
→∞ln = lim
→∞ln = ∞, so ∞
=1
diverges by part (a).
42. Let = 1
2 and = 1
. Then lim
→∞
= lim
→∞
1
= 0, but
diverges while
converges.
43. lim
→∞= lim
→∞
1, so we apply the Limit Comparison Test with = 1
. Since lim
→∞ 0we know that either both series converge or both series diverge, and we also know that ∞
=1
1
diverges [-series with = 1]. Therefore,
must be divergent.
44. First we observe that, by l’Hospital’s Rule, lim
→0
ln(1 + )
= lim
→0
1
1 + = 1. Also, if
converges, then lim
→∞= 0by Theorem 11.2.6. Therefore, lim
→∞
ln(1 + )
= lim
→0
ln(1 + )
= 1 0. We are given that
is convergent and 0.
Thus,
ln(1 + )is convergent by the Limit Comparison Test.
45. Yes. Since
is a convergent series with positive terms, lim
→∞= 0by Theorem 11.2.6, and
=
sin()is a series with positive terms (for large enough ). We have lim
→∞
= lim
→∞
sin()
= 1 0by Theorem 2.4.2 [ET Theorem 3.3.2]. Thus,
is also convergent by the Limit Comparison Test.
46. Yes. Since
converges, its terms approach 0 as → ∞, so for some integer , ≤ 1 for all ≥ . But then
∞
=1=−1
=1 +∞
=≤−1
=1 +∞
=. The first term is a finite sum, and the second term converges since∞
=1converges. So
converges by the Comparison Test.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
53. IfP an is a convergent series with positive terms, is it true thatP sin(an) is also convergent?
Solution:
142 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 40. (a) Since lim
→∞() = 0, there is a number 0 such that |− 0| 1 for all , and so since
and are positive. Thus, since
converges, so does
by the Comparison Test.
(b) (i) If = ln
3 and = 1
2, then lim
→∞
= lim
→∞
ln
= lim
→∞
ln
= limH
→∞
1
1 = 0, so ∞
=1
ln
3 converges by part (a).
(ii) If = ln
√ and = 1
, then lim
→∞
= lim
→∞
ln √
= lim
→∞
ln √
= limH
→∞
1
1(2√
) = lim
→∞
√2
= 0. Now
is a convergent geometric series with ratio = 1 [|| 1], so
converges by part (a).
41. (a) Since lim
→∞
= ∞, there is an integer such that
1whenever . (Take = 1 in Definition 11.1.5.) Then whenever and since
is divergent,
is also divergent by the Comparison Test.
(b) (i) If = 1
ln and = 1
for ≥ 2, then lim
→∞
= lim
→∞
ln = lim
→∞
ln
= limH
→∞
1
1 = lim
→∞ = ∞, so by part (a), ∞
=2
1
ln is divergent.
(ii) If = ln
and = 1
, then ∞
=1
is the divergent harmonic series and lim
→∞
= lim
→∞ln = lim
→∞ln = ∞, so ∞
=1
diverges by part (a).
42. Let = 1
2 and = 1
. Then lim
→∞
= lim
→∞
1
= 0, but
diverges while
converges.
43. lim
→∞= lim
→∞
1, so we apply the Limit Comparison Test with = 1
. Since lim
→∞ 0we know that either both series converge or both series diverge, and we also know that ∞
=1
1
diverges [-series with = 1]. Therefore,
must be divergent.
44. First we observe that, by l’Hospital’s Rule, lim
→0
ln(1 + )
= lim
→0
1
1 + = 1. Also, if
converges, then lim
→∞= 0by Theorem 11.2.6. Therefore, lim
→∞
ln(1 + )
= lim
→0
ln(1 + )
= 1 0. We are given that
is convergent and 0.
Thus,
ln(1 + )is convergent by the Limit Comparison Test.
45. Yes. Since
is a convergent series with positive terms, lim
→∞= 0by Theorem 11.2.6, and
=
sin()is a series with positive terms (for large enough ). We have lim
→∞
= lim
→∞
sin()
= 1 0by Theorem 2.4.2 [ET Theorem 3.3.2]. Thus,
is also convergent by the Limit Comparison Test.
46. Yes. Since
converges, its terms approach 0 as → ∞, so for some integer , ≤ 1 for all ≥ . But then
∞
=1=−1
=1 +∞
=≤−1
=1 +∞
=. The first term is a finite sum, and the second term converges since∞
=1converges. So
converges by the Comparison Test.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
55. LetP an andP bn be series with positive terms. Is each of the following statements true or false? If the statement is false, give an example that disproves the statement.
(a) IfP an andP bn are divergent, thenP anbn is divergent.
(b) IfP an converges andP bn diverges, thenP anbn diverges.
(c) IfP an andP bn are convergent, thenP anbn is convergent.
1
Solution:
SECTION 11.5 ALTERNATING SERIES AND ABSOLUTE CONVERGENCE ¤ 1089
51. lim
→∞= lim
→∞
1, so we apply the Limit Comparison Test with = 1
. Since lim
→∞ 0we know that either both series converge or both series diverge, and we also know that ∞
=1
1
diverges [series with = 1]. Therefore,
must be divergent.
52. First we observe that, by l’Hospital’s Rule, lim
→0
ln(1 + )
= lim
→0
1
1 + = 1. Also, if
converges, then lim
→∞= 0by Theorem 11.2.6. Therefore, lim
→∞
ln(1 + )
= lim
→0
ln(1 + )
= 1 0. We are given that
is convergent and 0.
Thus,
ln(1 + )is convergent by the Limit Comparison Test.
53. Yes. Since
is a convergent series with positive terms, lim
→∞= 0by Theorem 11.2.6, and
=
sin()is a
series with positive terms (for large enough ). We have lim
→∞
= lim
→∞
sin()
= 1 0by Theorem 3.3.5. Thus,
is also convergent by the Limit Comparison Test.
54. Since
converges, lim
→∞= 0, so there exists such that |− 0| 1 for all ⇒ 0 ≤ 1for all ⇒ 0 ≤ 2≤ . Since
converges, so does
2by the Comparison Test.
55. (a) False. The series
= 1
and
= 1
are both divergent, but
= 1
· 1
= 1
2 is convergent.
(b) False. The series
= 1
2 converges and
= 1
diverges, but
= 1
2 · 1
= 1
3 converges.
(c) True. Since
converges, for suffficiently large , 1 ⇒ . Since
converges, it follows by the Direct Comparison Test that
converges.
11.5 Alternating Series and Absolute Convergence
1. (a) An alternating series is a series whose terms are alternately positive and negative.
(b) An alternating series ∞
=1
= ∞
=1(−1)−1, where = ||, converges if 0 +1≤ for all and lim
→∞= 0.
(This is the Alternating Series Test.)
(c) The error involved in using the partial sum as an approximation to the total sum is the remainder = − and the size of the error is smaller than +1; that is, || ≤ +1. (This is the Alternating Series Estimation Theorem.)
2. 2 3−2
5+2 7−2
9+ 2
11 − · · · = ∞
=1(−1)+1 2
2 + 1. Now = 2
2 + 1 0, {} is decreasing, and lim→∞= 0, so the series converges by the Alternating Series Test.
° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
2
